At the end of $t$ years, the future value of an investment of $\$ 10,000$ in an account that pays $12 \%$ APR compounded monthly is $S=10,000\left(1+\frac{0.12}{12}\right)^{12 t}$ dç्. lars. Assuming no withdrawals or additional deposits, how long will it take for the investment to amount to $\$ 40,000$ ?

The investment will grow to $\$ 40,000$ in $\square$ years.
(Do not round until the final answer. Then round to two decimal places as needed.)

Step 1 ：Given values are: Future value of the investment (S) = $40,000, Initial investment (P) = $10,000, Annual interest rate (r) = 12%, Number of times interest is compounded per year (n) = 12.

Step 2 ：We need to find the time (t) it takes for the investment to grow to $40,000.

Step 3 ：We use the formula for the future value of an investment compounded monthly: \(S = P \left(1 + \frac{r}{n}\right)^{nt}\).

Step 4 ：Rearranging the formula to solve for t, we get \(t = \frac{\ln(\frac{S}{P})}{n \ln(1 + \frac{r}{n})}\).

Step 5 ：Substituting the given values into the formula, we get \(t = \frac{\ln(\frac{40000}{10000})}{12 \ln(1 + \frac{0.12}{12})}\).

Step 6 ：Calculating the above expression, we find that \(t \approx 11.61\).

Step 7 ：So, it will take approximately 11.61 years for the investment to grow to $40,000.

Step 8 ：Final Answer: The investment will grow to $40,000 in \(\boxed{11.61}\) years.